Problem: $\begin{aligned} y&=\sqrt{4x+1} \\\\ \dfrac{dy}{dx}&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $2\sqrt{4x+1}$ (Choice B) B $\dfrac{2}{\sqrt{4x+1}}$ (Choice C) C $\dfrac{2}{\sqrt{x}}$ (Choice D) D $\dfrac{1}{2\sqrt{4x+1}}$
Since $\sqrt{4x+1}$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $\underbrace{\sqrt{~\overbrace{4x+1}^{\text{inner}}~}}_{\text{outer}}$ So if $\sqrt{4x+1}=w(u(x))$, then: $\begin{aligned} {u(x)}&={4x+1} &&\text{inner function} \\\\ w(x)&=\sqrt{x}&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={4} \\\\ {w'(x)}&={\dfrac{1}{2\sqrt{x}}} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={\dfrac{1}{2\sqrt{({4x+1})}}} \cdot {4} \\\\ &=\dfrac{2}{\sqrt{4x+1}} \end{aligned}$